-0.000 000 000 010 068 461 037 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 068 461 037 33(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 010 068 461 037 33(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 010 068 461 037 33| = 0.000 000 000 010 068 461 037 33


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 010 068 461 037 33.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 010 068 461 037 33 × 2 = 0 + 0.000 000 000 020 136 922 074 66;
  • 2) 0.000 000 000 020 136 922 074 66 × 2 = 0 + 0.000 000 000 040 273 844 149 32;
  • 3) 0.000 000 000 040 273 844 149 32 × 2 = 0 + 0.000 000 000 080 547 688 298 64;
  • 4) 0.000 000 000 080 547 688 298 64 × 2 = 0 + 0.000 000 000 161 095 376 597 28;
  • 5) 0.000 000 000 161 095 376 597 28 × 2 = 0 + 0.000 000 000 322 190 753 194 56;
  • 6) 0.000 000 000 322 190 753 194 56 × 2 = 0 + 0.000 000 000 644 381 506 389 12;
  • 7) 0.000 000 000 644 381 506 389 12 × 2 = 0 + 0.000 000 001 288 763 012 778 24;
  • 8) 0.000 000 001 288 763 012 778 24 × 2 = 0 + 0.000 000 002 577 526 025 556 48;
  • 9) 0.000 000 002 577 526 025 556 48 × 2 = 0 + 0.000 000 005 155 052 051 112 96;
  • 10) 0.000 000 005 155 052 051 112 96 × 2 = 0 + 0.000 000 010 310 104 102 225 92;
  • 11) 0.000 000 010 310 104 102 225 92 × 2 = 0 + 0.000 000 020 620 208 204 451 84;
  • 12) 0.000 000 020 620 208 204 451 84 × 2 = 0 + 0.000 000 041 240 416 408 903 68;
  • 13) 0.000 000 041 240 416 408 903 68 × 2 = 0 + 0.000 000 082 480 832 817 807 36;
  • 14) 0.000 000 082 480 832 817 807 36 × 2 = 0 + 0.000 000 164 961 665 635 614 72;
  • 15) 0.000 000 164 961 665 635 614 72 × 2 = 0 + 0.000 000 329 923 331 271 229 44;
  • 16) 0.000 000 329 923 331 271 229 44 × 2 = 0 + 0.000 000 659 846 662 542 458 88;
  • 17) 0.000 000 659 846 662 542 458 88 × 2 = 0 + 0.000 001 319 693 325 084 917 76;
  • 18) 0.000 001 319 693 325 084 917 76 × 2 = 0 + 0.000 002 639 386 650 169 835 52;
  • 19) 0.000 002 639 386 650 169 835 52 × 2 = 0 + 0.000 005 278 773 300 339 671 04;
  • 20) 0.000 005 278 773 300 339 671 04 × 2 = 0 + 0.000 010 557 546 600 679 342 08;
  • 21) 0.000 010 557 546 600 679 342 08 × 2 = 0 + 0.000 021 115 093 201 358 684 16;
  • 22) 0.000 021 115 093 201 358 684 16 × 2 = 0 + 0.000 042 230 186 402 717 368 32;
  • 23) 0.000 042 230 186 402 717 368 32 × 2 = 0 + 0.000 084 460 372 805 434 736 64;
  • 24) 0.000 084 460 372 805 434 736 64 × 2 = 0 + 0.000 168 920 745 610 869 473 28;
  • 25) 0.000 168 920 745 610 869 473 28 × 2 = 0 + 0.000 337 841 491 221 738 946 56;
  • 26) 0.000 337 841 491 221 738 946 56 × 2 = 0 + 0.000 675 682 982 443 477 893 12;
  • 27) 0.000 675 682 982 443 477 893 12 × 2 = 0 + 0.001 351 365 964 886 955 786 24;
  • 28) 0.001 351 365 964 886 955 786 24 × 2 = 0 + 0.002 702 731 929 773 911 572 48;
  • 29) 0.002 702 731 929 773 911 572 48 × 2 = 0 + 0.005 405 463 859 547 823 144 96;
  • 30) 0.005 405 463 859 547 823 144 96 × 2 = 0 + 0.010 810 927 719 095 646 289 92;
  • 31) 0.010 810 927 719 095 646 289 92 × 2 = 0 + 0.021 621 855 438 191 292 579 84;
  • 32) 0.021 621 855 438 191 292 579 84 × 2 = 0 + 0.043 243 710 876 382 585 159 68;
  • 33) 0.043 243 710 876 382 585 159 68 × 2 = 0 + 0.086 487 421 752 765 170 319 36;
  • 34) 0.086 487 421 752 765 170 319 36 × 2 = 0 + 0.172 974 843 505 530 340 638 72;
  • 35) 0.172 974 843 505 530 340 638 72 × 2 = 0 + 0.345 949 687 011 060 681 277 44;
  • 36) 0.345 949 687 011 060 681 277 44 × 2 = 0 + 0.691 899 374 022 121 362 554 88;
  • 37) 0.691 899 374 022 121 362 554 88 × 2 = 1 + 0.383 798 748 044 242 725 109 76;
  • 38) 0.383 798 748 044 242 725 109 76 × 2 = 0 + 0.767 597 496 088 485 450 219 52;
  • 39) 0.767 597 496 088 485 450 219 52 × 2 = 1 + 0.535 194 992 176 970 900 439 04;
  • 40) 0.535 194 992 176 970 900 439 04 × 2 = 1 + 0.070 389 984 353 941 800 878 08;
  • 41) 0.070 389 984 353 941 800 878 08 × 2 = 0 + 0.140 779 968 707 883 601 756 16;
  • 42) 0.140 779 968 707 883 601 756 16 × 2 = 0 + 0.281 559 937 415 767 203 512 32;
  • 43) 0.281 559 937 415 767 203 512 32 × 2 = 0 + 0.563 119 874 831 534 407 024 64;
  • 44) 0.563 119 874 831 534 407 024 64 × 2 = 1 + 0.126 239 749 663 068 814 049 28;
  • 45) 0.126 239 749 663 068 814 049 28 × 2 = 0 + 0.252 479 499 326 137 628 098 56;
  • 46) 0.252 479 499 326 137 628 098 56 × 2 = 0 + 0.504 958 998 652 275 256 197 12;
  • 47) 0.504 958 998 652 275 256 197 12 × 2 = 1 + 0.009 917 997 304 550 512 394 24;
  • 48) 0.009 917 997 304 550 512 394 24 × 2 = 0 + 0.019 835 994 609 101 024 788 48;
  • 49) 0.019 835 994 609 101 024 788 48 × 2 = 0 + 0.039 671 989 218 202 049 576 96;
  • 50) 0.039 671 989 218 202 049 576 96 × 2 = 0 + 0.079 343 978 436 404 099 153 92;
  • 51) 0.079 343 978 436 404 099 153 92 × 2 = 0 + 0.158 687 956 872 808 198 307 84;
  • 52) 0.158 687 956 872 808 198 307 84 × 2 = 0 + 0.317 375 913 745 616 396 615 68;
  • 53) 0.317 375 913 745 616 396 615 68 × 2 = 0 + 0.634 751 827 491 232 793 231 36;
  • 54) 0.634 751 827 491 232 793 231 36 × 2 = 1 + 0.269 503 654 982 465 586 462 72;
  • 55) 0.269 503 654 982 465 586 462 72 × 2 = 0 + 0.539 007 309 964 931 172 925 44;
  • 56) 0.539 007 309 964 931 172 925 44 × 2 = 1 + 0.078 014 619 929 862 345 850 88;
  • 57) 0.078 014 619 929 862 345 850 88 × 2 = 0 + 0.156 029 239 859 724 691 701 76;
  • 58) 0.156 029 239 859 724 691 701 76 × 2 = 0 + 0.312 058 479 719 449 383 403 52;
  • 59) 0.312 058 479 719 449 383 403 52 × 2 = 0 + 0.624 116 959 438 898 766 807 04;
  • 60) 0.624 116 959 438 898 766 807 04 × 2 = 1 + 0.248 233 918 877 797 533 614 08;
  • 61) 0.248 233 918 877 797 533 614 08 × 2 = 0 + 0.496 467 837 755 595 067 228 16;
  • 62) 0.496 467 837 755 595 067 228 16 × 2 = 0 + 0.992 935 675 511 190 134 456 32;
  • 63) 0.992 935 675 511 190 134 456 32 × 2 = 1 + 0.985 871 351 022 380 268 912 64;
  • 64) 0.985 871 351 022 380 268 912 64 × 2 = 1 + 0.971 742 702 044 760 537 825 28;
  • 65) 0.971 742 702 044 760 537 825 28 × 2 = 1 + 0.943 485 404 089 521 075 650 56;
  • 66) 0.943 485 404 089 521 075 650 56 × 2 = 1 + 0.886 970 808 179 042 151 301 12;
  • 67) 0.886 970 808 179 042 151 301 12 × 2 = 1 + 0.773 941 616 358 084 302 602 24;
  • 68) 0.773 941 616 358 084 302 602 24 × 2 = 1 + 0.547 883 232 716 168 605 204 48;
  • 69) 0.547 883 232 716 168 605 204 48 × 2 = 1 + 0.095 766 465 432 337 210 408 96;
  • 70) 0.095 766 465 432 337 210 408 96 × 2 = 0 + 0.191 532 930 864 674 420 817 92;
  • 71) 0.191 532 930 864 674 420 817 92 × 2 = 0 + 0.383 065 861 729 348 841 635 84;
  • 72) 0.383 065 861 729 348 841 635 84 × 2 = 0 + 0.766 131 723 458 697 683 271 68;
  • 73) 0.766 131 723 458 697 683 271 68 × 2 = 1 + 0.532 263 446 917 395 366 543 36;
  • 74) 0.532 263 446 917 395 366 543 36 × 2 = 1 + 0.064 526 893 834 790 733 086 72;
  • 75) 0.064 526 893 834 790 733 086 72 × 2 = 0 + 0.129 053 787 669 581 466 173 44;
  • 76) 0.129 053 787 669 581 466 173 44 × 2 = 0 + 0.258 107 575 339 162 932 346 88;
  • 77) 0.258 107 575 339 162 932 346 88 × 2 = 0 + 0.516 215 150 678 325 864 693 76;
  • 78) 0.516 215 150 678 325 864 693 76 × 2 = 1 + 0.032 430 301 356 651 729 387 52;
  • 79) 0.032 430 301 356 651 729 387 52 × 2 = 0 + 0.064 860 602 713 303 458 775 04;
  • 80) 0.064 860 602 713 303 458 775 04 × 2 = 0 + 0.129 721 205 426 606 917 550 08;
  • 81) 0.129 721 205 426 606 917 550 08 × 2 = 0 + 0.259 442 410 853 213 835 100 16;
  • 82) 0.259 442 410 853 213 835 100 16 × 2 = 0 + 0.518 884 821 706 427 670 200 32;
  • 83) 0.518 884 821 706 427 670 200 32 × 2 = 1 + 0.037 769 643 412 855 340 400 64;
  • 84) 0.037 769 643 412 855 340 400 64 × 2 = 0 + 0.075 539 286 825 710 680 801 28;
  • 85) 0.075 539 286 825 710 680 801 28 × 2 = 0 + 0.151 078 573 651 421 361 602 56;
  • 86) 0.151 078 573 651 421 361 602 56 × 2 = 0 + 0.302 157 147 302 842 723 205 12;
  • 87) 0.302 157 147 302 842 723 205 12 × 2 = 0 + 0.604 314 294 605 685 446 410 24;
  • 88) 0.604 314 294 605 685 446 410 24 × 2 = 1 + 0.208 628 589 211 370 892 820 48;
  • 89) 0.208 628 589 211 370 892 820 48 × 2 = 0 + 0.417 257 178 422 741 785 640 96;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 010 068 461 037 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1000 1100 0100 0010 0001 0(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 037 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1000 1100 0100 0010 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 010 068 461 037 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1000 1100 0100 0010 0001 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1000 1100 0100 0010 0001 0(2) × 20 =

1.0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010(2) × 2-37


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -37

Mantissa (not normalized):
1.0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

986(10) =

011 1101 1010(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010 =

0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1101 1010

Mantissa (52 bits) =
0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010


Decimal number -0.000 000 000 010 068 461 037 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 0111 1111 0001 1000 1000 0100 0010

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100